Now, assume that it holds for n = m, (where m ≥ 1), so you assume the following is true.

[itex]\displaystyle\sum_{i=0}^{2m-1}(3i+1)= m(6m-1)\,.[/itex]​

With that assumption, you now need to show that this formulation is true for n = m+1. In this case, 2n-1 = 2(m+1)-1 = 2m+1, and n(6n-1)= (m+1)(6m+5)= 6m2+11m+5 . In other words, show that the following can be derived from the above.

Thank you SammyS for your reply your hint lead me to the proper proof. I was having a hard time figuring out why it would be (3(2m)+1) and (3(2m+1)+1) instead of just (3(m+1)+1) ect.... But i think i get it now.